Step-by-step explanation:
CH 4 ( g ) + 2 O 2 ( g ) ⟶ CO 2 ( g ) + 2 H 2 O ( l )
enthalpy of formation of CO 2 ( g )
− 4 × bond energy of C H
4 × bond energy of C − H
enthalpy of combustion of CH 4
Answer:
enthalpy of combustion of CH 4: This is because we usually calculate the enthalpy change of combustion from enthalpies of formation. This is given as;
ΔHoc = ∑ΔH∘f(products) − ∑ΔH∘f(reactants)
CH 4 ( g ) ⟶ C ( g ) + 4 H ( g )
1. enthalpy of combustion of CH4
2. enthalpy of formation of C(g)
3. 4× bond energy of C–H
4. –4× bond energy of C–H
Answer:
4× bond energy of C–H: Bond breaking is an endothermic process. So it cannot be option 4, enthalpy of formation of a pure element is 0 KJ/mol. This is not a combustion reaction, so options 1 and 2 are aslo wrong.
c) C(graphite) + 2H2(g) ------------ CH4(g)
1. enthalpy of combustion of CH4
2. enthalpy of formation of CH4(g)
3. 4× bond energy of C–H
4. –4× bond energy of C–H
Answer:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. The correct option is; enthalpy of formation of CH4(g)
d) C(g) + 4H(g) ------------------ CH4(g)
1. enthalpy of combustion of C
2. enthalpy of formation of CH4(g)
3. 4× bond energy of C–H
4. –4× bond energy of C–H
Answer:
–4× bond energy of C–H; bond formation is always exothermic, it cannot be option 3. This is not a combustion reaction so it cannot be option 1. enthalpy of formation of a pure element is 0 KJ/mol, so it cannot be option 2.